Question

The component shown below is made of AISI 4340 Oil Quenched at 800oC and Temper at 595oC 1 hour ( σ y 139ksi; σ ut: 147ks) tubular steel. The component, a closed end tube with an OD1.25in and ID 3/4in, is subjected to the load (see arrow) that varies between 3F and -IF Determine the force F for infinite life. Given: The component has been cold rolled, operates at room temperature and needs a reliability of 99.0%. Consider all stress concentrations equals to onc 12 in 4 in I 2 in 6 in.

Answer 1

Psine OD= 1.25in ID 3/4 in Fto-F gin Z/ ㅓan-115): 22-62。 Shifting Psíns to B" will result in transverse load Psinθ and a torque of magnitud (Psine x 6in) .: Torque = PSin (2262") x 6in = 2.307 P and this torque remains constant throughou the Span AB Psin(2262") transverse laad will induce a marlmum bending rmoment oF magnitude Psin (2 62) x 12 in? at ffrxed end {For this B-M c', lies in neutral Rbre B.M-Psin(2262.) x 12 = 4.615 P Shifting Pcose to B will result in a compressive load of maghitude cos and a constant moment of magnitude Poosex 6 Compressive Load- P cos(22-62) = o. 923P Moment = Ros (22620) x 6in = 5538 p {for this A', lies in neutral fibre? For Design consi deration .. The components Subjected to B-M= 5,588 P Torque 2 307P Compressive loads 0-923 p x 125 2. : Critical Bending stress M5.538 P x 64 二 33,182 P

Critical Torsional stress I. Ymax 23071 x 125 (十254-0-754) 32. 6.911P Critical compressive load0:923P 밖 1,115 P Tx (1-252-0152 4 .: Normal stress-ơr(33182 +1175) p Torsional shear stress 6-9I1 P Dmax= 34.357 x 3F 34,397 P 34357x -F-34 357F min 103.071 F min6.911 x -F-6911 F 20-733 F Tmean (20133 6911)F6911F meanl03 071 -34.35734 357 F 2. Ơamp 20133+6.911 F2 131 22 F 103-071.+34 357F- Tamp 68, 714 F 2. 72-765F For steels Corrected Endurance limit actor Ka -87 as mwthing mentfoned about machined surface is cold rolled 1 Kw oeg: Size-factor dependingon dea meter Kc 0.814 corresponding to 99% relrability Se 087 x1x0.8 14-х 73-5 = 52,0513 Kei

For infinite life Assuming Fos-d Max value ofGp = Se a m 2.765 F52 0813 F 52.0513 = 0.7153 Kips 72 765 F 715-3 1b Hence for any value of F less than 7I53 lb we woill obtain Hen ce for any inilit infinite life for the component